Ran in:

0 个投票

Ran in:
Hi, is it possible to avoid the loop? Thanks.
x=(1:3:r);
Unrecognized function or variable 'r'.
[r,c]=size(Tr);
E1=zeros(r,c);
for i=1:c
E1(x,i)=E(Tr(r,i),i);
end
>> size(Tr)
ans =
21 6

5 个评论
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DGM
DGM 2023-7-29
What is z? What is Tr?
What is the value of r before it's redefined? Is it the same?
pipin
pipin 2023-7-29
excuse me.. z=r...i change it in code
Tr is a matrix
pipin
pipin 2023-7-29
[r,c]=size(Tr);
x=(1:3:r);
E1=zeros(r,c);
for i=1:c
E1(x,i)=E(Tr(r,i),i);
end
Image Analyst
Image Analyst 2023-7-29
Not seeing it. Again, what is Tr and E (not E1, but E)?
Does your for loop even work? What is the desired output?
If you have any more questions, then attach your data and missing code to read it in after you read this:
TUTORIAL: How to ask a question (on Answers) and get a fast answer
pipin
pipin 2023-7-29
thanks your help I made some mistakes while creating the code because it's a bit laborious. Here's the correct version
%input:
row=10;
columns=5;
step=3;
%%*************
E = round(rand(row,columns)*10)
Tr = randi(size(E,1),row,size(E,2))
[r,c]=size(Tr);
x=(1:step:r);
% Your method
E1=zeros(r,c);
for i=1:c
Tr(x,i)
E(Tr(x,i),i)
E1(x,i)=E(Tr(x,i),i);
end
E1

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 采纳的回答

Bruno Luong
Bruno Luong 2023-7-30
编辑：Bruno Luong 2023-7-30

1 个投票

Just to pull out the right answer (my previous answer applied on a WRONG calculation specified by OP)
% vectorized method
c = size(Tr,2);
m = size(E,1);
x = 1:step:m;
E1 = zeros(size(Tr));
E1(x,:) = E(Tr(x,:) + (0:c-1)*m);

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pipin
pipin 2023-7-30
编辑：pipin 2023-7-30
hello.. why do you say that the previous answer was wrong? on the isequal test it was correct
(what is OP?)
Bruno Luong
Bruno Luong 2023-7-30
编辑：Bruno Luong 2023-7-30
OP is Original Poster, it's you who wrote : "excuse me.. z=r...i change it in code" ... remember?
My answer (not the comments below it) is wrong because the question is wrong.

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更多回答（1 个）

Bruno Luong
Bruno Luong 2023-7-29
编辑：Bruno Luong 2023-7-29
Ran in:

1 个投票

Ran in:
If you want obscure code by avoiding for-loop
E = rand(10,5)
E = 10×5
0.4433 0.5126 0.5907 0.7564 0.9740 0.9908 0.4916 0.9634 0.8613 0.0549 0.9769 0.2085 0.4765 0.2052 0.7497 0.0450 0.8816 0.9243 0.6475 0.1040 0.8076 0.1900 0.4109 0.6581 0.2044 0.4685 0.1936 0.7925 0.7647 0.4082 0.9868 0.9336 0.2478 0.0403 0.7413 0.1741 0.5068 0.1629 0.0827 0.0434 0.1120 0.7868 0.0827 0.4696 0.7350 0.4545 0.5871 0.0055 0.1273 0.2868
Tr = randi(size(E,1),9,size(E,2))
Tr = 9×5
9 10 5 2 4 9 10 9 9 2 1 10 4 6 2 2 8 2 3 4 8 2 7 9 8 5 7 2 4 5 5 5 7 9 5 5 10 10 4 3 9 8 3 3 7
[r,c]=size(Tr);
x=(1:3:r);
[r,c]=size(Tr);
x=(1:3:r);
% Your method
E1=zeros(r,c);
for i=1:c
E1(x,i)=E(Tr(r,i),i);
end
E1
E1 = 9×5
0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0
% vectorized method
E1=zeros(r,c);
i=1:c;
E1(x(:) + (i-1)*r) = repmat(E(Tr(r,i) + (i-1)*size(E,1)),length(x),1);
E1
E1 = 9×5
0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0

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Bruno Luong
Bruno Luong 2023-7-29
PS: I have a feeling that is NOT what you expect to get.
pipin
pipin 2023-7-29
thanks your help I made some mistakes while creating the code because it's a bit laborious. Here's the correct version
%input:
row=10;
columns=5;
step=3;
%%*************
E = round(rand(row,columns)*10)
Tr = randi(size(E,1),row,size(E,2))
[r,c]=size(Tr);
x=(1:step:r);
% Your method
E1=zeros(r,c);
for i=1:c
Tr(x,i)
E(Tr(x,i),i)
E1(x,i)=E(Tr(x,i),i);
end
E1
Bruno Luong
Bruno Luong 2023-7-29
Ran in:
Ran in:
"I made some mistakes while creating the code because it's a bit laborious. "
I knew it, not it is not because laborious, it's because you are simply not serious on what you are doing.
row=10;
columns=5;
step=3;
%%*************
E = round(rand(row,columns)*10)
E = 10×5
3 10 7 6 1 1 6 1 1 1 8 2 4 3 5 7 9 6 6 5 9 8 1 7 5 4 10 3 6 1 7 9 8 5 3 2 3 8 4 1 10 1 3 0 2 2 5 2 8 2
Tr = randi(size(E,1),row,size(E,2))
Tr = 10×5
3 7 9 10 2 4 3 2 5 8 4 10 9 3 10 10 3 5 7 6 5 6 1 10 7 3 5 4 6 1 2 10 8 5 1 8 1 8 1 2 7 7 10 1 10 1 9 1 7 2
[r,c]=size(Tr);
x=(1:step:r);
% Your method
E1=zeros(r,c);
for i=1:c
E1(x,i)=E(Tr(x,i),i);
end
E1
E1 = 10×5
8 9 3 8 1 0 0 0 0 0 0 0 0 0 0 2 2 1 5 1 0 0 0 0 0 0 0 0 0 0 1 5 8 7 1 0 0 0 0 0 0 0 0 0 0 3 1 7 5 1
% vectorized method
E1=zeros(r,c);
i=1:c;
E1(x(:) + (i-1)*r) = E(Tr(x,i) + (i-1)*size(E,1));
E1
E1 = 10×5
8 9 3 8 1 0 0 0 0 0 0 0 0 0 0 2 2 1 5 1 0 0 0 0 0 0 0 0 0 0 1 5 8 7 1 0 0 0 0 0 0 0 0 0 0 3 1 7 5 1
pipin
pipin 2023-7-29
:D
very good
pipin
pipin 2023-7-29
编辑：pipin 2023-7-29
i try it with
row=500000;
columns=500;
step=30;
Elapsed time is 20.148787 seconds. (mYcODE)
Elapsed time is 18.746811 seconds. (VECTORIZED CODE)
Why do you think vectorization doesn't improve execution time?
i use:
Processore Intel(R) Core(TM) i5-7200U CPU @ 2.50GHz 2.71 GHz
RAM installata 8,00 GB (7,88 GB utilizzabile)
Tipo sistema Sistema operativo a 64 bit, processore basato su x64
Bruno Luong
Bruno Luong 2023-7-29
编辑：Bruno Luong 2023-7-29
Ran in:
Ran in:
Why do you think it does?
This is the timining of the Online server, your PC does not have anough memory IMO. Each array E, E1, and Tr take 2 Gb, so a total of 6 Gb of free RAM is needed.
row=500000;
columns=500;
step=300;
%%*************
E = round(rand(row,columns)*10);
Tr = randi(size(E,1),row,size(E,2));
tic
[r,c]=size(Tr);
x=(1:step:r);
% Your method
E1=zeros(r,c);
for i=1:c
E1(x,i)=E(Tr(x,i),i);
end
toc
Elapsed time is 0.607406 seconds.
tic
% vectorized method II
[r,c]=size(Tr);
x=(1:step:size(E,1));
E1=zeros(r,c);
E1(x,:) = E(Tr(x,:) + (0:c-1)*size(E,1));
toc
Elapsed time is 0.633170 seconds.
pipin
pipin 2023-7-29
编辑：pipin 2023-7-29
20 sec (my method) vs 18 sec (vectorized) are similar
often with vectorization I see advantages of 1:10 as speed compared to loops
likely because this pc has few cores and is not benefited by parallel computing
pipin
pipin 2023-7-29
ok thank
Bruno Luong
Bruno Luong 2023-7-29
编辑：Bruno Luong 2023-7-29
"often with vectorization I see advantages of 1:10 as speed compared to loops"
Your information is surely NOT up-to-date and too generic to have any real value.
pipin
pipin 2023-8-1
编辑：pipin 2023-8-1
hi, i'm using your method vectorized
% vectorized method II
[r,c]=size(Tr);
x=(1:step:size(E,1));
E1=zeros(r,c);
E1(x,:) = E(Tr(x,:) + (0:c-1)*size(E,1));
toc
I have seen that Tr can also assume the value zero which would lead to an error in the function (you cannot search for a zero index)
is it possible to do the search of the indexes excluding if the value is zero?
i solve it:
[r,c]=size(MinTrade_Tab_Short);
a=MinTrade_Tab_Short+ (0:c-1)*r;
a2=a+(~a); %tolgo lo zero (per evitare errore riga sucessiva)
E1 = E(a2);
E2=E1.*(a>0);
it's good for you?
Bruno Luong
Bruno Luong 2023-8-2
编辑：Bruno Luong 2023-8-2
No your solution is NOT good as it only catches Tr == 0 at column 0. For other column a2 is not 0 so you won't correct the overflowed row indexing.
Bruno Luong
Bruno Luong 2023-8-2
编辑：Bruno Luong 2023-8-2
Code to handle the case Tr == 0
c = size(Tr,2);
m = size(E,1);
x = 1:step:m;
E1 = zeros(size(Tr));
Trx = Tr(x,:);
iE = Trx + m*(0:c-1);
tmp = E(max(iE,1));
tmp(Trx == 0) = 0;
E1(x,:) = tmp;
it does not handle still the case Tr > m or Tr < 0.
If you havve data that are not valid for proper indexing it will be a mess to deal with.
pipin
pipin 2023-8-2
Correct..thank you!

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